How to prove that $(a+b-ab)^n+(1-a^n)(1-b^n) \geq n $ for $a,b\in [0,1]$ and $n\in\mathbb{N}$?

Question

Let $a,b\in [0,1]$ and $n\in\mathbb{N}$. Prove the following inequality:
$$(a+b-ab)^n+(1-a^n)(1-b^n) \geq n $$

I thought on using M Induction: Assuming that the inequality holds for $n=k,$ $k\in\mathbb{N}$ let's prove that:
$(a+b-ab)^{k+1}+(1-a^{k+1})(1-b^{k+1})\geq k+1 $

I tried to rearrange this one a little bit (so I could use the assumption) like this:
$(a+b-ab)^k(a+b-ab)+(1-a^ka)(1-b^kb)=$
$=(a+b-ab)^k+(a+b-ab-1)(a+b-ab)^k+(1-a^k)(1-b^k)-(1-b^k)(a-1)a^k-(1-a^k)(b-1)b^k+a^kb^k(a-1)(b-1)\geq$
$\geq k+(a+b-ab-1)(a+b-ab)^k-(1-b^k)(a-1)a^k-(1-a^k)(b-1)b^k+a^kb^k(a-1)(b-1)$

But I don't think I did anything useful here...

Answer 1

As user2345215 has said, it is false for $a=b=1$. It is also false for $a=b=0$ if $n>1$.

For $a=b=1/2$, we have $$\left(\frac34\right)^n+\left(\frac14\right)^n<\left(\frac34+\frac14\right)^n=1$$